ceaglex/VisualTTS_Chem · Datasets at Hugging Face

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fCGDtkmkMgU-009\|So silver, when it's reduced, can form oxidized copper.
fCGDtkmkMgU-012\|Silver metal will plate out on to copper.
fCGDtkmkMgU-014\|Now, you could also make gold plate out if you had the correct combination of ions.
fCGDtkmkMgU-015\|So if you said silver or gold in this case, you are correct.
B-HgZ3G95oA-000\|Let's look at the equilibrium situation between liquid water and gaseous water at 25 degrees C.
B-HgZ3G95oA-001\|What can we determine about the free energy difference and K?
B-HgZ3G95oA-002\|Now, here I'm talking about the free energy without the standard sign.
B-HgZ3G95oA-003\|So the free energy difference under the situation that exists right now.
B-HgZ3G95oA-012\|So delta G without the standard state is 0.
B-HgZ3G95oA-013\|It's the free energy difference between these actual concentrations at equilibrium.
B-HgZ3G95oA-014\|So at equilibrium you may remember that the partial pressure of gaseous water is around 0.03 of an atmosphere.
B-HgZ3G95oA-015\|So the free energy difference between gaseous water at 0.03 of an atmosphere and liquid water at 25 degrees C is 0.
B-HgZ3G95oA-016\|K is the equilibrium constant for that value, which is 0.03 of an atmosphere.
B-HgZ3G95oA-017\|So K is less than 1, 0.03, and delta G is 0.
B-HgZ3G95oA-018\|The correct answer here is B.
KT_N4_kKHnQ-000\|Gases are an ideal system to look at energy changes, because gases change their energy in direct proportion to temperature changes.
KT_N4_kKHnQ-001\|Remember, the energy change of a gas is 3/2 RT for one mole, 3/2 nRT for a given number of moles.
KT_N4_kKHnQ-002\|But if the temperature changes, the energy changes.
KT_N4_kKHnQ-003\|So an energy change always corresponds to a temperature change, so the gas is kind of transparent that way.
KT_N4_kKHnQ-004\|It can't hide an energy change because the temperature always changes.
KT_N4_kKHnQ-005\|So when work is done on a gas, that's an energy change and could be reflected in a temperature change.
KT_N4_kKHnQ-006\|If heat is absorbed by gas, that's an energy change, could be reflected in the temperature change.
KT_N4_kKHnQ-009\|So I have to keep track of both the heat and the work to manage energy changes.
KT_N4_kKHnQ-010\|Let's look at the work.
KT_N4_kKHnQ-011\|If I do work for a gas, I push a cylinder up in a gas, I change the height of this cylinder.
KT_N4_kKHnQ-012\|Work is just force times distance, or in this case, force times the height.
KT_N4_kKHnQ-016\|If I expand, the system is also doing the work, and that should take some of the system's energy to do that work.
KT_N4_kKHnQ-020\|If a gas is compressed, then I do work on the gas.
KT_N4_kKHnQ-023\|I'm the surroundings, the system is the gas, I compress it, the energy of the system tends to go up, work is positive in those circumstances.
KT_N4_kKHnQ-024\|So that's how work, the change in volume, and external pressure are involved in calculating energy changes for gases.
V-FBHIdXk20-000\|Let's look at the hydrogen-like energy systems.
V-FBHIdXk20-001\|Now when I say hydrogen-like, I mean the one-electron systems.
V-FBHIdXk20-002\|Hydrogen, a proton and an electron, that's a complete hydrogen atom.
V-FBHIdXk20-004\|Or you could go lithium plus 2.
V-FBHIdXk20-005\|As you have one-electron systems, they all behave just like hydrogen atoms.
V-FBHIdXk20-006\|So let's look at those.
V-FBHIdXk20-007\|Which ion does the energy-level diagram X correspond to?
V-FBHIdXk20-008\|So here's hydrogen and near it, some unknown ion, X.
V-FBHIdXk20-019\|Here we have hydrogen and some other ion.
V-FBHIdXk20-020\|Now, what lines up here?
V-FBHIdXk20-021\|We know that the energies go as minus Z squared over n squared.
V-FBHIdXk20-022\|And we see in this system that level 2 in the unknown system lines up with level 1 in the hydrogen system.
V-FBHIdXk20-023\|And that energy 1 for this system is 1 squared over 1 squared times i infinity.
V-FBHIdXk20-024\|That's the hydrogen atom.
V-FBHIdXk20-025\|For this one, it has to be something over something to still give me minus R infinity.
V-FBHIdXk20-026\|The value of Z is undetermined here because we don't know which ion it is.
V-FBHIdXk20-027\|But the value of n is 2.
V-FBHIdXk20-031\|So if it has two protons-- I don't care what else it has in it-- it is a helium nucleus.
V-FBHIdXk20-032\|So a helium plus is the ion that would have this configuration on its electronic energy scale.
V-FBHIdXk20-033\|Z will be 2 for helium.
z76W5579a5Y-000\|The decomposition of hydrogen peroxide, H2O2, is exothermic.
z76W5579a5Y-001\|Now, if you look carefully at this reaction, what can you determine about the structure of H2O2 simply knowing that that reaction is exothermic?
z76W5579a5Y-012\|So energetically, that doesn't tell us anything about whether this reaction is endothermic or exothermic.
z76W5579a5Y-013\|What about the oxygen bond?
z76W5579a5Y-014\|On this side, I'm forming oxygen gas, so O2 double-bonded.
z76W5579a5Y-015\|And I form half a mole of those.
z76W5579a5Y-016\|On this side, I have to break a mole of oxygen-oxygen bonds.
z76W5579a5Y-017\|The question is, are they double bonds or single bonds?
z76W5579a5Y-019\|And we know it's not endothermic, so we can already rule out C and B.
z76W5579a5Y-020\|Now, what about if it's a single bond?
z76W5579a5Y-022\|That is, the double bonds have to be more than twice as energetic to be stable as the single bonds.
z76W5579a5Y-025\|So that will give us a net exothermic reaction, and the correct answer here is A, the single-bonded structure.
1iUgRsNvJkA-000\|The isomerisation of propanol to isopropanol is endothermic.
1iUgRsNvJkA-001\|Now what does that tell you about burning propanol and isopropanol?
1iUgRsNvJkA-002\|Which burns cooler?
1iUgRsNvJkA-010\|Propanol and isopropanol both burn in oxygen to form carbon dioxide in water.
1iUgRsNvJkA-011\|And they are the same chemical formula, but different structure.
1iUgRsNvJkA-012\|They're structural isomers.
1iUgRsNvJkA-013\|Does that structurally isomerisation cause a different enthalpy of burning?
1iUgRsNvJkA-018\|But they go to the same products.
1iUgRsNvJkA-019\|So isopropanol burning to give the same products this same state is a greater drop in enthalpy.
1iUgRsNvJkA-020\|So there's a greater release in energy on the burning of isopropanol.
1iUgRsNvJkA-021\|So propanol burns cooler.
1iUgRsNvJkA-022\|The correct answer here is A.
qpmTRXIJhFo-000\|When we talk about ionization energy, we're talking about how easy is it to remove an electron from an atom?
qpmTRXIJhFo-001\|Well, what holds the electron about the atom?
qpmTRXIJhFo-002\|It's the positive charge on the nucleus.
qpmTRXIJhFo-003\|So the more positive charge that an electron can see, the more difficult it is to ionize.
qpmTRXIJhFo-004\|And by contrast, the more of that charge that's shielded, the easier it is to remove that electron.
qpmTRXIJhFo-005\|And easier to remove means higher energy.
qpmTRXIJhFo-006\|Those energy levels go up closer to the 0 state.
qpmTRXIJhFo-007\|Remember, the 0 state for the nucleus and the electron is the ionized state.
qpmTRXIJhFo-008\|So you get closer to the ionized state, the more charge you can shield from outer electrons.
qpmTRXIJhFo-009\|Let's look at that.
qpmTRXIJhFo-010\|So if we have electrons, hydrogen and helium plus, those are both one electron systems, and they're very easy to understand.
qpmTRXIJhFo-011\|Their ionization energy will just be z squared-- the charge on the nucleus-- over n-- whichever orbital the electron happens to be in.
qpmTRXIJhFo-013\|And that's identically four times that.
qpmTRXIJhFo-014\|They're one electron systems.
qpmTRXIJhFo-015\|The charge on the nucleus is plus 2 for helium.
qpmTRXIJhFo-016\|Plus 2 squared is 4, so you have four times the ionization energy for helium, as opposed to hydrogen.
qpmTRXIJhFo-017\|Now, what if you start to put some electrons in the helium?
qpmTRXIJhFo-018\|We can look at helium in the 1s2 state.
qpmTRXIJhFo-020\|If it's sufficient to ionize, the electrons will be ejected.
qpmTRXIJhFo-021\|And if you put even more energy in, that electron goes-- that energy goes into kinetic energy of the electron.
qpmTRXIJhFo-022\|So let's go ahead and look at helium in the 1s2 state.
qpmTRXIJhFo-023\|Now the electric ionization energy is given by the effective charge squared over n squared.
qpmTRXIJhFo-026\|Well, the answer is no.
qpmTRXIJhFo-027\|Two s electrons don't shield each other fully.
qpmTRXIJhFo-029\|So the s electrons-- there's a little bit of shielding going on, but it's not a full nuclear charge that shielded.
qpmTRXIJhFo-030\|How about if one of those electrons is promoted, is excited, out to a 2 state, the 2p, for instance?

fCGDtkmkMgU-009|So silver, when it's reduced, can form oxidized copper.

fCGDtkmkMgU-012|Silver metal will plate out on to copper.

fCGDtkmkMgU-014|Now, you could also make gold plate out if you had the correct combination of ions.

fCGDtkmkMgU-015|So if you said silver or gold in this case, you are correct.

B-HgZ3G95oA-000|Let's look at the equilibrium situation between liquid water and gaseous water at 25 degrees C.

B-HgZ3G95oA-001|What can we determine about the free energy difference and K?

B-HgZ3G95oA-002|Now, here I'm talking about the free energy without the standard sign.

B-HgZ3G95oA-003|So the free energy difference under the situation that exists right now.

B-HgZ3G95oA-012|So delta G without the standard state is 0.

B-HgZ3G95oA-013|It's the free energy difference between these actual concentrations at equilibrium.

B-HgZ3G95oA-014|So at equilibrium you may remember that the partial pressure of gaseous water is around 0.03 of an atmosphere.

B-HgZ3G95oA-015|So the free energy difference between gaseous water at 0.03 of an atmosphere and liquid water at 25 degrees C is 0.

B-HgZ3G95oA-016|K is the equilibrium constant for that value, which is 0.03 of an atmosphere.

B-HgZ3G95oA-017|So K is less than 1, 0.03, and delta G is 0.

B-HgZ3G95oA-018|The correct answer here is B.

KT_N4_kKHnQ-000|Gases are an ideal system to look at energy changes, because gases change their energy in direct proportion to temperature changes.

KT_N4_kKHnQ-001|Remember, the energy change of a gas is 3/2 RT for one mole, 3/2 nRT for a given number of moles.

KT_N4_kKHnQ-002|But if the temperature changes, the energy changes.

KT_N4_kKHnQ-003|So an energy change always corresponds to a temperature change, so the gas is kind of transparent that way.

KT_N4_kKHnQ-004|It can't hide an energy change because the temperature always changes.

KT_N4_kKHnQ-005|So when work is done on a gas, that's an energy change and could be reflected in a temperature change.

KT_N4_kKHnQ-006|If heat is absorbed by gas, that's an energy change, could be reflected in the temperature change.

KT_N4_kKHnQ-009|So I have to keep track of both the heat and the work to manage energy changes.

KT_N4_kKHnQ-010|Let's look at the work.

KT_N4_kKHnQ-011|If I do work for a gas, I push a cylinder up in a gas, I change the height of this cylinder.

KT_N4_kKHnQ-012|Work is just force times distance, or in this case, force times the height.

KT_N4_kKHnQ-016|If I expand, the system is also doing the work, and that should take some of the system's energy to do that work.

KT_N4_kKHnQ-020|If a gas is compressed, then I do work on the gas.

KT_N4_kKHnQ-023|I'm the surroundings, the system is the gas, I compress it, the energy of the system tends to go up, work is positive in those circumstances.

KT_N4_kKHnQ-024|So that's how work, the change in volume, and external pressure are involved in calculating energy changes for gases.

V-FBHIdXk20-000|Let's look at the hydrogen-like energy systems.

V-FBHIdXk20-001|Now when I say hydrogen-like, I mean the one-electron systems.

V-FBHIdXk20-002|Hydrogen, a proton and an electron, that's a complete hydrogen atom.

V-FBHIdXk20-004|Or you could go lithium plus 2.

V-FBHIdXk20-005|As you have one-electron systems, they all behave just like hydrogen atoms.

V-FBHIdXk20-006|So let's look at those.

V-FBHIdXk20-007|Which ion does the energy-level diagram X correspond to?

V-FBHIdXk20-008|So here's hydrogen and near it, some unknown ion, X.

V-FBHIdXk20-019|Here we have hydrogen and some other ion.

V-FBHIdXk20-020|Now, what lines up here?

V-FBHIdXk20-021|We know that the energies go as minus Z squared over n squared.

V-FBHIdXk20-022|And we see in this system that level 2 in the unknown system lines up with level 1 in the hydrogen system.

V-FBHIdXk20-023|And that energy 1 for this system is 1 squared over 1 squared times i infinity.

V-FBHIdXk20-024|That's the hydrogen atom.

V-FBHIdXk20-025|For this one, it has to be something over something to still give me minus R infinity.

V-FBHIdXk20-026|The value of Z is undetermined here because we don't know which ion it is.

V-FBHIdXk20-027|But the value of n is 2.

V-FBHIdXk20-031|So if it has two protons-- I don't care what else it has in it-- it is a helium nucleus.

V-FBHIdXk20-032|So a helium plus is the ion that would have this configuration on its electronic energy scale.

V-FBHIdXk20-033|Z will be 2 for helium.

z76W5579a5Y-000|The decomposition of hydrogen peroxide, H2O2, is exothermic.

z76W5579a5Y-001|Now, if you look carefully at this reaction, what can you determine about the structure of H2O2 simply knowing that that reaction is exothermic?

z76W5579a5Y-012|So energetically, that doesn't tell us anything about whether this reaction is endothermic or exothermic.

z76W5579a5Y-013|What about the oxygen bond?

z76W5579a5Y-014|On this side, I'm forming oxygen gas, so O2 double-bonded.

z76W5579a5Y-015|And I form half a mole of those.

z76W5579a5Y-016|On this side, I have to break a mole of oxygen-oxygen bonds.

z76W5579a5Y-017|The question is, are they double bonds or single bonds?

z76W5579a5Y-019|And we know it's not endothermic, so we can already rule out C and B.

z76W5579a5Y-020|Now, what about if it's a single bond?

z76W5579a5Y-022|That is, the double bonds have to be more than twice as energetic to be stable as the single bonds.

z76W5579a5Y-025|So that will give us a net exothermic reaction, and the correct answer here is A, the single-bonded structure.

1iUgRsNvJkA-000|The isomerisation of propanol to isopropanol is endothermic.

1iUgRsNvJkA-001|Now what does that tell you about burning propanol and isopropanol?

1iUgRsNvJkA-002|Which burns cooler?

1iUgRsNvJkA-010|Propanol and isopropanol both burn in oxygen to form carbon dioxide in water.

1iUgRsNvJkA-011|And they are the same chemical formula, but different structure.

1iUgRsNvJkA-012|They're structural isomers.

1iUgRsNvJkA-013|Does that structurally isomerisation cause a different enthalpy of burning?

1iUgRsNvJkA-018|But they go to the same products.

1iUgRsNvJkA-019|So isopropanol burning to give the same products this same state is a greater drop in enthalpy.

1iUgRsNvJkA-020|So there's a greater release in energy on the burning of isopropanol.

1iUgRsNvJkA-021|So propanol burns cooler.

1iUgRsNvJkA-022|The correct answer here is A.

qpmTRXIJhFo-000|When we talk about ionization energy, we're talking about how easy is it to remove an electron from an atom?

qpmTRXIJhFo-001|Well, what holds the electron about the atom?

qpmTRXIJhFo-002|It's the positive charge on the nucleus.

qpmTRXIJhFo-003|So the more positive charge that an electron can see, the more difficult it is to ionize.

qpmTRXIJhFo-004|And by contrast, the more of that charge that's shielded, the easier it is to remove that electron.

qpmTRXIJhFo-005|And easier to remove means higher energy.

qpmTRXIJhFo-006|Those energy levels go up closer to the 0 state.

qpmTRXIJhFo-007|Remember, the 0 state for the nucleus and the electron is the ionized state.

qpmTRXIJhFo-008|So you get closer to the ionized state, the more charge you can shield from outer electrons.

qpmTRXIJhFo-009|Let's look at that.

qpmTRXIJhFo-010|So if we have electrons, hydrogen and helium plus, those are both one electron systems, and they're very easy to understand.

qpmTRXIJhFo-011|Their ionization energy will just be z squared-- the charge on the nucleus-- over n-- whichever orbital the electron happens to be in.

qpmTRXIJhFo-013|And that's identically four times that.

qpmTRXIJhFo-014|They're one electron systems.

qpmTRXIJhFo-015|The charge on the nucleus is plus 2 for helium.

qpmTRXIJhFo-016|Plus 2 squared is 4, so you have four times the ionization energy for helium, as opposed to hydrogen.

qpmTRXIJhFo-017|Now, what if you start to put some electrons in the helium?

qpmTRXIJhFo-018|We can look at helium in the 1s2 state.

qpmTRXIJhFo-020|If it's sufficient to ionize, the electrons will be ejected.

qpmTRXIJhFo-021|And if you put even more energy in, that electron goes-- that energy goes into kinetic energy of the electron.

qpmTRXIJhFo-022|So let's go ahead and look at helium in the 1s2 state.

qpmTRXIJhFo-023|Now the electric ionization energy is given by the effective charge squared over n squared.

qpmTRXIJhFo-026|Well, the answer is no.

qpmTRXIJhFo-027|Two s electrons don't shield each other fully.

qpmTRXIJhFo-029|So the s electrons-- there's a little bit of shielding going on, but it's not a full nuclear charge that shielded.

qpmTRXIJhFo-030|How about if one of those electrons is promoted, is excited, out to a 2 state, the 2p, for instance?